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Is there a general way to solve equations like the one below? $$ a = b^{y} - c^{y} $$ Where b > 1, c < 1 and a > b? For example...

$$ 5 = 1.1^{y} - 0.1^{y} $$

Moreover, would the problem be any easier to solve if we could choose what b and c were...so long as b>1 and c<1?

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    Well, if you are fine with numerical methods, use the [Newton-Raphson method](https://en.m.wikipedia.org/wiki/Newton's_method). – 2017-02-19

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If $\frac{\ln b}{\ln c}\in \mathbb{Q}$, then there are $p,q\in\mathbb{Z}$ with $b^p = c^q$ and $q\geq 0$. Let $d:=c^{\frac{1}{p}}=b^{\frac{1}{q}}$. Then $a=b^y-c^y$ becomes $a=d^{yq}-d^{yp}$. If we substitute $t=d^y$, we get $$a=t^q-t^p.$$ If $p$ and $q$ are positive, this is a polynomial of degree $\max\{p,q\}$. If $p$ is negative, we multiply the equation with $t^{-p}$ to get a polynomial of degree $q-p$. We can now try to find the roots of the polynomial and then get the solution by "unrolling" the substitution(s). $$ y=\log_d t = \frac{\ln t}{\ln d} $$ If $\frac{\ln b}{\ln c}\not\in \mathbb{Q}$, then a numerical method must be used.

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    Why use ln $ln$ $t$ instead of $log_x$ $t$? – 2017-02-20
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    @Blake No special reason. When I used $\frac{\ln b}{\ln c}$ instead of $\log_c b$, I only wanted to emphasize the fact that $b$ and $c$ play similar roles. $\log_c b$ looks as if there was something special about $c$. And if you can choose the base of the logarithm, then $\ln$ is the most natural choice. When I used $\frac{\ln t}{\ln d}$, I only wanted to mention how to calculate a logarithm with given base. – 2017-02-20
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    I thought about that later, and came to the same conclusion. I do agree, that the OP would wonder about 'c' and how to find c, although it is good to understand that c is any number – 2017-02-20